Problem: $f(x)=\dfrac{\sqrt{x}}{2}$ Find the second degree Taylor polynomial, centered at $x=4$, of $f$. Choose 1 answer: Choose 1 answer: (Choice A) A $1+\frac{1}{8}x-\frac{1}{128}x^2$ (Choice B) B $1+\frac{1}{8}x+\frac{1}{128}x^2$ (Choice C) C $1+\frac{1}{8}(x-4)-\frac{1}{64}(x-4)^2$ (Choice D) D $1+\frac{1}{8}(x-4)-\frac{1}{128}(x-4)^2$ (Choice E) E $1+\frac{1}{8}(x+4)-\frac{1}{128}(x+4)^2$
Explanation: First, find the first two derivatives of $~f(x)=\dfrac{\sqrt{x}}{2}=\dfrac{1}{2}x^{1/2}\,$. ${f}\,^\prime(x)=\dfrac{1}{4}x^{-1/2}=\frac{1}{4x^{1/2}}$ ${f}\,^{\prime\prime}(x)=-\dfrac{1}{8}x^{-3/2}=-\dfrac{1}{8x^{3/2}}$ Then let $~x=4~$ in the original function and these derivatives to get the coefficients for $~{{T}_{2}}\left( x \right)\,$, the second degree Taylor polynomial for $~f\left( x \right)\,$. $ f(4)=1\,; {f}'(4)=\frac{1}{8}\,; {f}''(4)=-\frac{1}{64}$ Use these coefficients in the equation of the Taylor polynomial of degree $~2\,$. ${{T}_{3}}\left( x \right)=f(4)+{f}\,^\prime(4)(x-4)+\frac{{f}\,^{\prime\prime}(4){{(x-4)}^{2}}}{2!}$ Hence, ${{T}_{2}}\left( x \right)=1+\frac{1}{8}(x-4)-\frac{1}{128}(x-4)^2$